2024 AMC 12B Problem 25

Question

Pablo will decorate each of

6

identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the

12

decisions he must make. After the paint dries, he will place the

6

balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events “the ball Frida selects is red” and “the ball Frida selects is striped” may or may not be independent, depending on the outcome of Pablo’s coin flips. The probability that these two events are independent can be written as

\dfrac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m?

(Recall that two events

A

and

B

are independent if

P(A \text{ and } B) = P(A)\cdot P(B).

)

243

245

247

249

251

Answer: A

Concepts:independent events casework

Difficulty rating: 2510

Small hint:

Let the six balls include

k

that are red-and-striped, with

R

red and

S

striped total. Independence for Frida’s uniform pick means

\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6}

Big hint:

So the condition is

6k = RS.

Count assignments of

6

balls to the

4

equally likely types satisfying this, over

4^6

total, using multinomial coefficients

Solution:

Each ball is independently one of four equally likely types: red-striped, red-dotted, blue-striped, blue-dotted. Suppose among the

6

balls there are

k

red-striped, with

R

red and

S

striped in total. For Frida’s uniform pick,

P(\text{red}) = \tfrac{R}{6},

P(\text{striped}) = \tfrac{S}{6},

and

P(\text{red and striped}) = \tfrac{k}{6}.

Independence means

\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6},

i.e.

6k = RS.

First count assignments with

R\in\{0,6\}

or

S\in\{0,6\}.

Each of these four conditions leaves

2^6=64

choices for the other attribute, and the four assignments at their pairwise intersections have been counted twice. Their union therefore contributes

4\cdot64-4=252.

For

1\le R,S\le5,

the condition that

RS

be divisible by

6

leaves only

(R,S)=(2,3),(3,2)

or

(3,4),(4,3).

In each case

k=\frac{RS}{6},

and the four type counts are a permutation of

(1,1,2,2).

Thus each pair contributes

\dfrac{6!}{1!1!2!2!}=180

assignments. The favorable count is therefore

252+4\cdot180=972.

Out of

4^6=4096

equally likely assignments, the probability is

\dfrac{972}{4096}=\dfrac{243}{1024},

so

m=243.

Thus, the correct answer is A.